Titration problem?

INTERNATIONAL JUNIOR SCIENCE OLYMPIAD , FIFTH EDITION , GYEONGNAM , SOUTH KOREA

A weak acid HA is titrated with a standard base ( NaOH solution) and the following titration curve is obtained.

For a weak acid , the dissociation constant , Ka , is defined as:

HA + H2O <=> #A^- + H3O^+#

Ka= [A-][H3O+]/[HA]

The realtionship between Ka and pH is :

-lgKa=pH - lg{[A-]/[HA]}

and pH = -lg[H3O+]

Using the information provided determine the dissociation constant ,Ka , for the weak acid HA.

a) #10^{-4,2}#

b)#10^{-4,5}#

c)#10^{-7,8}#

d)#10^{-11,5}#

2 Answers
Jun 21, 2018

Answer:

#"Ka"= 10^(-4.5)#

Explanation:

The #"pKa"# of a monoprotic weak acid is the same as the #"pH"# halfway to its equivalence point. Here's how this shortcut works:

Sodium hydroxide #"NaOH"# neutralizes the weak acid #"HA"# to produce #"NaA"# by the equation

#"NaOH" + "HA" to "NaA" + "H"_2"O" color(grey)(" balanced.")#

One formula unit of sodium hydroxide reacts with one #"HA"# molecule to produce one formula unit of #"NaA"#. Thus

#n("NaA")=n("HA","reacted")=n("NaOH", "added")#

Appearantly
#n("HA", "reacted")+ n("HA", "remaining") = n("HA", "initial")#

Substituting #n("HA","reacted")# in the second equation with #n("NaA")# gives

#n("HA","remaining") = n("HA", "initial")- n("NaA")#

#n("NaA") = n("NaOH") = n("HA", "initial")# at the equivalence point where #V("NaOH", "added") = 22 color(white)(l) "ml"# for this question;

Therefore halfway to that point at the titration midpoint at which #11 color(white)(l) "ml"# of #"NaOH"# has been added, #n("NaA") = n("NaOH") = color(navy)(1/2)n("HA", "initial")#

#n("HA","remaining") = n("HA", "initial")- n("NaA")#
#color(white)(n("HA","remaining")) =color(navy)(1/2)n("HA", "initial")#
#color(white)(n("HA","remaining") )=n("NaA")#

Additionally, given that

#"NaA" (aq) color(purple) to "Na"^(+) + "A"^-#

#"NaA"#, a salt, completely disassociates as a strong electrolyte,

#n("A"^-)=n("NaA")= n("HA")#.

#"Ka"# is defined as the equilibrium constant of the reaction

#"HA" (aq) rightleftharpoons "H"^+ (aq) + "A"^(-) (aq)#

#"Ka"="K"_"eq"=(["H"^+] color(red)(cancel(color(black)(["A"^(-)]))))/(color(red)(cancel(color(black)(["HA"]))))=["H"^+]#

given that #n("A"^-)=n("NaA")= n("HA")# as previously stated.

Therefore

#"Ka"=["H"^+]=10^(-"pH")=10^(-4.5)#

Reference
"Titration Fundamentals", Chemistry LibreText, https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_Fundamentals

See also
"Titration of a Weak Base with a Strong Acid", https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Base_with_a_Strong_Acid

Jun 21, 2018

Answer:

#K_a = 10^(-4.5)#

Explanation:

As an alternative approach, you can use the fact that at the half-equivalence point, exactly half of the weak acid has been neutralized by the strong base.

The chemical equation that describes this neutralization reaction looks like this

#"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

At the half-equivalence point, you add enough moles of strong base, which I've represented here by #"OH"^(-)#, to consume half of the initial number of moles of the weak acid.

Since the reaction consumes the weak acid and produces the conjugate base #"A"^(-)# in a #1:1# mole ratio, you can say that at the half-equivalence point, you have

#["HA"] = ["A"^(-)]#

Now, notice that at the equivalence point, #"22 mL"# of the strong base have been added to the weak acid solution. This tells you that it takes #"22 mL"# of the strong base to consume all the moles of the weak acid present in the initial solution.

So if it takes #"22 mL"# of the strong base to consume all the moles of the weak acid, it follows that if you add #"11 mL"# of the strong base, you will consume exactly half of the moles of the weak acid.

This means that the half-equivalence point corresponds to #"11 mL"# of the strong base added and a #"pH"# of #4.5#.

The question provides you with the equation

#-"lg"(K_a) = "pH" - "lg"( (["A"^(-)])/(["HA"]))#

which is a version of the Henderson - Hasselbalch equation that can be used to calculate the #"pH"# of a weak acid - conjugate base buffer.

Plug in the #"pH"# of the solution at the half-equivalence point and the fact that #["HA"] = ["A"^(-)]# to get

#- "lg"(K_a)= 4.5 - "lg"(1)#

This gets you

#"lg"(K_a) = - 4.5#

which implies

#color(darkgreen)(ul(color(black)(K_a = 10^(-4.5))))#

See this answer for a similar question.

SIDE NOTE: The notation used here is based on the fact that

#"lg"(x) = log_(10)(x)#

See here for more info on that.